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At the finals of the "Gaana Sunao" contest, which was conducted all over India, five contestants - C1, C2, C3, C4 and C5 - participated. Before the announcement of the results, six mischievous persons - Azad, Bose, Chand, Dev, Ehsaan and Fardeen - managed to get hold of the result sheet, which contained the scores of each of the five participants. Each of the six persons then decided to announce the sum of the final scores of exactly four contestants. So, Azad, Bose, Chand, Dev, Ehsaan and Fardeen announced their sums as 220, 260, 230, 240, 210 and 250 points respectively. However, one of them made a mistake in adding the scores. Also, the organizers of the contest decided to award an amount of Rs.10,000 for each point that a contestant scored. The score of each contestant is an integral value.

Question :- What is the highest possible amount that any of the five contestants can be eligible for?

(1) Rs.7 lakh

(2) Rs.7.5 lakh

(3) Rs.8 lakh

(4) Rs.8.5 lakh

(5) Rs.9 lakh

My attempt :-

I assumed that scores of the 5 participants are as follows $C_1<=C_2<=C_3<=C_4<=C_5$

There are only 5 combinations of 4 scores that could have been possible:-

$1) C_1, C_2, C_3, C_4$

$2) C_1, C_2, C_3, C_5$

$3) C_1, C_2, C_4, C_5$

$4) C_1, C_3, C_4, C_5$

$5) C_2, C_3, C_4, C_5$

Now since there are 6 unique scores given and of which one is incorrect , I deduced that the incorrect score would be a duplicate of one of the other 5 scores

Now to find the highest possible amount one can get I assumed that the 210 score would be true, now this score would have been achieved by the 4 lowest scorers or (Total score of all the 5 participants)-(Largest score)

$Total-Largest=210$

$Largest=Total-210$

Now from this equation Largest score can be anything as it will be dependent on the total score

How to proceed from here ?

Fin27
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  • When you add the five correct sums you get four times the total of all five scores. Once you decide to throw out the $210$, the scores are $C_1, C_1+10, C_1+20, C_1+30, C_1+40$. You can solve for $C_1$. I don't think it is obvious to throw out the $210$ as throwing out a middle score increases the dispersion and may increase the highest score. Note that the sum of the five sums you keep must be a multiple of $40$. – Ross Millikan Apr 09 '23 at 14:36
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    "The score of each contestant is an integral value." This information helps to narrow down who made a mistake in adding the scores. – peterwhy Apr 09 '23 at 14:37
  • @RossMillikan how did you get $C_1, C_1+10, C_1+20, C_1+30, C_1+40$ and what are you equating this to? – Fin27 Apr 09 '23 at 21:47
  • Once you throw out $210$ each sum of scores is $10$ more than the previous one. That comes from replacing one score with another, so the new score is $10$ more than the old one. If you add these up you get $4C_1+100$, which is the sum of the five sums of scores. – Ross Millikan Apr 09 '23 at 23:28

1 Answers1

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Okay, first define: $$x_i = \sum_{j\neq i} C_j$$ Observe that $x_i$ are the five possible sums (the correct ones).

Notice that: $$Total = x_i + C_i$$ for any $i \in \{1,2,3,4,5\}$. We get: $\color{green}{Total} = \color{blue}{x_1+C_1}=\color{blue}{x_2+C_2} = \color{blue}{x_3+C_3}=\color{blue}{x_4+C_4}=\color{blue}{x_5+C_5}$ Sum the green part $5$ times, and equate it to the sum of the blue things.
$$5\cdot Total = \sum_{i=1}^5 (x_i+C_i) = \sum_{i=1}^5 x_i+\sum_{i=1}^5 C_i = \sum_{i=1}^5 x_i+Total$$ So, $$Total = \frac 14 \sum_{i=1}^5 x_i$$

Since the total must be an integer, it follows that $\sum_{i=1}^5 x_i$ is divisible by $4$. Checking the $6$ sums modulo $4$, we find that one out of $210,230,250$ must be discarded.

Now, your formula becomes $$Largest = \frac 14 \sum_{i=1}^5 x_i - x_k = \frac 14 \left(\sum_{i=1}^5 x_i - 4x_k\right)$$ where $x_k$ is the smallest $x_i$. Clearly, $\sum_{i=1}^5 x_i$ will include $220,240,260$. So, discard one value from $210,230,250$ one at a time, then find the sum of the remaining two, and subtract $4$ times the minimum of the $5$ elements.

Case 1 ($210$ is discarded):
Here, $x_k = 220$. $$230+250 - 4\cdot 220 = \color{blue}{-400}$$

Case 2 ($230$ is discarded):
Here, $x_k = 210$. $$210+250 = 4\cdot 210 = \color{blue}{-380}$$

Case 3 ($250$ is discarded):
Here, $x_k = 210$. $$230+210-4\cdot 210 = \color{blue}{-400}$$

Clearly, we get a maximum when $230$ is discarded. Thus $$Largest = \frac 14(210+220+240+250+260 - 4\cdot 210) = 85$$ Thus, the answer is $\boxed{850000}$.

D S
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